AQA A-Level Chemistry Paper 1, 2022: Question 2

12 marks · Medium difficulty · State/Explain/Numerical

Define relative atomic mass, calculate an unknown isotopic mass, explain isotope chemical reactivity, perform a TOF mass spectrometry flight time calculation, and describe ion detection.

Practise this question

Question

Question 2 consists of five parts about rhenium (atomic number 75). Part 02.1 asks for the definition of relative atomic mass for 2 marks. Part 02.2 gives Table 2 showing isotope 185 with abundance 10, and an unknown isotope with abundance 17, and asks to calculate the unknown isotopic mass given an Ar of 186.3 (2 marks). Part 02.3 asks why rhenium isotopes have identical chemical properties (1 mark). Part 02.4 asks to calculate the flight time in seconds for a 185Re+ ion with kinetic energy 1.153 × 10^-13 J travelling through a 1.450 m flight tube using KE = 1/2 m v^2 and the Avogadro constant (5 marks). Part 02.5 asks how relative abundance of 185Re+ is determined in a TOF mass spectrometer (2 marks).
Question text

02 Rhenium has an atomic number of 75

02.1 Define the term relative atomic mass.

[2 marks]

02.2 The relative atomic mass of a sample of rhenium is 186.3

Table 2 shows information about the two isotopes of rhenium in this sample.

Table 2

Relative isotopic mass Relative abundance

185 10

To be calculated 17

Calculate the relative isotopic mass of the other rhenium isotope.

Show your working.

[2 marks]

Relative isotopic mass

02.3 State why the isotopes of rhenium have the same chemical properties.

[1 mark]

A sample of rhenium is ionised by electron impact in a time of flight (TOF) mass

spectrometer.

02.4 A 185Re+ ion with a kinetic energy of 1.153 × 10−13 J travels through a

1.450 m flight tube.

*06* 1 2

The kinetic energy of the ion is given by the equation KE = mv

where

m = mass / kg

v = speed / m s–1

KE = kinetic energy / J

Calculate the time, in seconds, for the ion to reach the detector.

The Avogadro constant, L = 6.022 × 1023 mol−1

[5 marks]

Time s

02.5 State how the relative abundance of 185Re+ is determined in a TOF mass

spectrometer.

[2 marks]

Mark scheme

Show the mark scheme Mark scheme for Question 2 in five sections. 02.1: average mass of 1 atom of an element compared to 1/12 mass of one atom of 12C (2 marks). 02.2: 186.3 = (185 × 10 + X × 17)/27, gives isotopic mass = 187 (2 marks). 02.3: same electron configuration/number of electrons (1 mark). 02.4: M1 mass of one ion = 3.072 × 10^-25 kg; M2 v = d/t; M3 v^2 = 2KE/m; M4 v = 8.66 × 10^5 m/s; M5 t = 1.67 × 10^-6 s (5 marks). 02.5: ions hit detector plate and gain electrons, producing a current proportional to abundance (2 marks).

Question Answers Additional Comments/Guidelines Mark

average/mean mass of 1 atom (of an element) M1 = top line 1

1/12 mass of one atom of 12C M2 = bottom line 1

or AO1

average/mean mass of atoms of an element if moles and atoms/isotopes mixed max = 1

1/12 mass of one atom of 12C

or

average/mean mass of atoms of an element ×12

mass of one atom of 12C

02.1 or

(average) mass of one mole of atoms

1/12 mass of one mole of 12C

or

(weighted) average mass of all the isotopes

1/12 mass of one atom of 12C

or

average mass of an atom/isotope (compared to C−12) on a

scale in which an atom of C−12 has a mass of 12

M1 186.3 = (185 × 10) + (X × 17) correct expression 1

02.2

M2 (relative isotopic mass) = 187(.1) 1

AO2

– A-LEVEL CHEMISTRY – –

same electron configuration allow same number of electrons 1

allow same electron structure AO1

02.3 ignore same number of protons

ignore different number of neutrons

do not accept–Asame number of neutrons-LEVEL CHEMISTRY– –

185 185 -25

M1 mass Re = 23 = 3.072 x 10 (kg) calculate mass in kg 1

6.02 × 10 × 1000

d

M2 v = recall of v = d/t 1

t

2 2KE 11

M3 v = or 7.5(0) × 10 rearrangement to get v2 1

m

02.4

2KE 5 –13

M4 v = √ or 8.66 × 10 √2 × 1.153 × 10 1

m allow

M1

1.45 –6 1.45

M5 t = 5 = 1.67 × 10 (s) M5 t = 1

8.66 × 10 M4

AO115

allow 1.67 × 10-6 to 1.68 × 10–6 (s)

– A-LEVEL CHEMISTRY – – AO2

alternative method:

185 185 -25

M1 mass Re = 23 = 3.072 x 10 (kg) calculate mass in kg 1

16 6.02 × 10 × 1000

d md2

M2 v = or KE = 2 recall of v = d/t 1

t 2t

md2

02.4 M3 t2 = rearrangement to get t2 1

2KE

m md2 3.072 × 10–25

√ √ M1

M4 t = d√ or or –13 allow √ –13 1

2KE 2KE 2 × 1.153 × 10 2 × 1.153 × 10

M5 t = 1.67 × 10–6 (s) allow 1.67 × 10–6 to 1.68 × 10–6 (s) 1

AO1

AO2

– A-LEVEL CHEMISTRY – –

at the detector/(negative) plate the ions/Re+ gain an electron 1

(relative) abundance depends on the size of the current 1

AO1

02.5

alternative answer

M1 ion knocks out an electron into electron

multiplier

M2 signal from electron multiplier proportional to

number of ions

How to answer it

Rhenium Isotopes and TOF Mass Spectrometry

📋 What This Question Tests

Core Specification Knowledge & Skills:

  • Fundamental Definitions: Precise standard definition of relative atomic mass (Ar) referencing carbon-12.
  • Isotopic Calculations: Rearranging weighted average abundance equations to determine unknown isotopic mass.
  • Chemical Reactivity & Subatomic Structure: Explaining why isotopes share identical chemical properties based on electronic structure.
  • Quantitative TOF Mass Spectrometry: Multi-step mathematical calculations involving kinetic energy ( KE = ½mv² ), velocity ( v = d/t ), Avogadro's constant, and crucial mass unit conversions ( g → kg ).
  • Detector Physics: Explaining the production of electrical signals and how ion abundance is quantified at the detector plate.
Question 02.1

Definition of Relative Atomic Mass

AQA Specification 3.1.1.1: Fundamental Particles • 2 Marks

✅ Correct Answer (Mark Scheme)

A rigorous two-part definition is required:

  • Mark 1 (Numerator): The average (or mean ) mass of one atom of an element
  • Mark 2 (Denominator): Divided by 1/12 the mass of one atom of ¹²C (carbon-12)

Alternative: The weighted average mass of an atom of an element compared to 1/12 the mass of a carbon-12 atom.

🧠 Exam Technique & Common Errors

  • ❌ Missing "Average": Writing simply "mass of an atom" instantly loses Mark 1. You must state average or mean mass.
  • ❌ Inconsistent Units: Never mix moles and atoms (e.g. "average mass of 1 mole compared to 1/12 mass of one atom" caps your score at 1 mark).
  • Tip: Always explicitly write the isotope symbol ¹²C or "carbon-12"; simply writing "carbon" is insufficient.
Examiner Insight: This standard definition is a frequent recall question. Top students score 2/2 easily by memorising the exact fraction phrase: "The average mass of an atom of an element / (1/12 the mass of an atom of carbon-12)".
Question 02.2

Calculating Unknown Relative Isotopic Mass

AQA Specification 3.1.1.2: Mass Spectrometry • 2 Marks

📐 Step-by-Step Calculation

  1. Total abundance calculation:
    Total abundance = 10 + 17 = 27
  2. Set up the weighted average formula (Mark 1):
    186.3 = [(185 × 10) + (X × 17)] / 27
  3. Rearrange to solve for X (Mark 2):
    186.3 × 27 = 1850 + 17X
    5030.1 = 1850 + 17X
    17X = 3180.1
    X = 187.06... → 187 (or 187.1)

❌ Common Errors to Avoid

  • Dividing by 100 instead of 27: The question gives relative abundances, not percentages. Always sum the given ratio values: 10 + 17 = 27 .
  • Unrealistic Answers: Rhenium's Ar is 186.3, and one isotope is 185. The other isotope must be heavier than 186.3. If you get a value like 184 or a huge number, check your algebra.
  • Rounding too early: Retain full calculator figures during intermediate steps.
Mark Breakdown: [M1] Correct algebraic expression set equal to 186.3 • [M2] Final answer of 187 (or 187.1).
Question 02.3

Chemical Properties of Isotopes

AQA Specification 3.1.1.1: Fundamental Particles • 1 Mark

✅ Correct Answer

They have the same electronic configuration (or same number of electrons / same electron structure).

💡 Key Chemistry Principle

Chemical reactions involve the loss, gain, or sharing of outer-shell electrons. Neutrons carry no charge and reside solely in the nucleus, affecting only physical properties (e.g. mass, density, rate of diffusion), not chemical reactivity.

  • Allowed: "Same number of electrons" or "same electronic arrangement".
  • Not allowed: Stating only "same number of protons" or "different number of neutrons" without mentioning electrons.
Examiner Insight: A classic 1-mark question. Students often write "they are the same element" or focus on protons. Chemical reactions are determined by electrons.
Question 02.4

TOF Flight Time Calculation

AQA Specification 3.1.1.2: Time of Flight Calculations • 5 Marks

📐 Comprehensive 5-Mark Step-by-Step Solution

  1. Calculate the mass of a single ¹⁸⁵Re⁺ ion in kilograms (Mark 1):
    First find mass of 1 mole in kg: 185 g mol⁻¹ = 185 × 10⁻³ kg mol⁻¹
    Divide by the Avogadro constant:
    m = 185 / (6.022 × 10²³ × 1000) = 3.07207 × 10⁻²⁵ kg
  2. Recall and link velocity to flight tube length and time (Mark 2):
    v = d / t
  3. Rearrange kinetic energy equation for v² (Mark 3):
    KE = ½mv²  ⟹  v² = 2KE / m
    v² = (2 × 1.153 × 10⁻¹³) / (3.07207 × 10⁻²⁵) = 7.5063 × 10¹¹ m² s⁻²
  4. Calculate velocity v (Mark 4):
    v = √(7.5063 × 10¹¹) = 8.6639 × 10⁵ m s⁻¹
  5. Calculate time t to reach the detector (Mark 5):
    t = d / v = 1.450 / (8.6639 × 10⁵) = 1.67 × 10⁻⁶ s  (allow 1.67 × 10⁻⁶ to 1.68 × 10⁻⁶ s)

🧠 Alternative Direct Substitution Method

Combine formulas before calculating:

t = d × √(m / 2KE)
t = 1.450 × √[(3.072 × 10⁻²⁵) / (2 × 1.153 × 10⁻¹³)]
t = 1.67 × 10⁻⁶ s

Both approaches earn full marks if intermediate steps or clear working are shown.

❌ Critical Calculation Traps

  • Forgetting to divide by 1000: The mass must be in kg because Joules are kg m² s⁻² . Using grams gives an answer out by a factor of √1000 ≈ 31.6.
  • Using wrong mass number: The question specifically states a ¹⁸⁵Re⁺ ion. Do not use the average atomic mass (186.3)!
  • Forgetting the square root: Students frequently find v² and plug it straight into t = d/v .
  • Units: Time in TOF spectrometry is exceptionally short (typically microsecond scale: 10⁻⁵ to 10⁻⁷ s ). A time of several seconds indicates a calculation error.
Question 02.5

Detection & Abundance in TOF Spectrometry

AQA Specification 3.1.1.2: Detection Stage • 2 Marks

✅ Correct Answer (Mark Scheme)

  • Mark 1 (Discharge): At the detector (negative plate), the positive ions (¹⁸⁵Re⁺) gain an electron (accept electron / are discharged).
  • Mark 2 (Signal): An electrical current is produced; the (relative) abundance is proportional to the size of the current.

🧠 Examiner Commentary & Key Keywords

Two distinct ideas are required for the 2 marks:

  1. Physical mechanism: Positive ions collide with the negatively charged detector plate and receive electrons to neutralise them.
  2. Signal generation: This transfer of charge creates a flow of electrons (a current). The greater the number of ions hitting the plate, the larger the current.

Alternative accepted: Ion knocks out an electron into an electron multiplier (M1), and the signal/current is proportional to the number of ions (M2).

Key takeaway: Never simply say "it is detected by a computer". The chemistry involves electron transfer causing an electric current proportional to abundance.

Topics

Physical Chemistry · 3.1.1 Atomic Structure · 3.1.2 Amount of Substance

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 1, 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.